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Home » Olympiad Section » Number Theory » Number Theory Unsolved Problems
 
divisor of a^2+b^2
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kimnimalar
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#1
divisor of a^2+b^2
if gcd(a;b)=1 then prove that a^2+b^2 has not divisor of the form 4k+3

PostPosted: Mon Jun 09, 2008 3:03 am  Back to top 
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aviateurpilot
Yang-Mills Theory
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#2
for p prime such that p|a^2+b^2 we have (ab^{-1})^2\equiv -1[p] then p\equiv 1[4]
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PostPosted: Mon Jun 09, 2008 10:12 am  Back to top 
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Armenia
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#3
If a^2+b^2 has a divisor of the form 4k+3
then there is a prime number p=4l+3, that a^2+b^2 \vdots p
From this we can say that a \vdots p and b \vdots p, or (a,b)\neq1.

PostPosted: Tue Jun 10, 2008 3:49 am  Back to top 
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kimnimalar
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#4
aviateurpilot wrote:
for p prime such that p|a^2 + b^2 we have (ab^{ - 1})^2\equiv - 1[p] then p\equiv 1[4]



what this mean (ab^(-1))^2=(-1) [p] ?

I can't understand it?

PostPosted: Sat Jun 14, 2008 5:21 am  Back to top 
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aviateurpilot
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#5
kimnimalar wrote:
aviateurpilot wrote:
for p prime such that p|a^2 + b^2 we have (ab^{ - 1})^2\equiv - 1[p] then p\equiv 1[4]



what this mean (ab^(-1))^2=(-1) [p] ?

I can't understand it?

b^{-1} it's just notation,
we now that if p\not|b, \exists u such that ub\equiv 1[mod\ p] (bezout) we note u=b^{-1}.
if p|a^2+b^2 then it's evident that p\not|b and a^{2}\equiv -b^2[mod\ p]
we take x=au : x^{2}\equiv a^2u^{2}\equiv -b^2u^2\equiv -1 (mod\ p) (here u=b^{-1} !! ok)
so if p=4h+3 then x^{p-1}=(x^{2})^{2h+1}\equiv (-1)^{2h+1}\equiv -1[p] (-impossible, by th.ferma)
then p\equiv 1[4]
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PostPosted: Sat Jun 14, 2008 9:05 am  Back to top 
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